# A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

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The countable chain conditon

A topological space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of discussion, if $X$ has the CCC, we also say that $X$ is a CCC space or X is CCC. A space $X$ is separable if there exists a countable subset $A$ of $X$ such that $A$ is dense in $X$ (meaning that if $U$ is a nonempty open subset of $X$, then $U \cap A \ne \varnothing$).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

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The strategy

Here’s the strategy for finding CCC and not separable.

The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

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Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, then consider $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, or consider $X^{2^c}$ where $X$ is your favorite separable space.

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Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the $\Sigma$-products. To make this clear, let’s focus on a specific example. Consider $X=\left\{0,1 \right\}^{c}$ where $c$ is the cardinality of continuum. Consider the following subspace.

$\Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}$

The subspace $\Sigma(\left\{0,1 \right\}^{c})$ is dense in $X$, thus has CCC. It is straightforward to verify that $\Sigma(\left\{0,1 \right\}^{c})$ is not separable.

To implement this example, find any space $X$ which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the $\Sigma$-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The $\Sigma$-product has the countable chain condition since it is a dense subspace of the CCC space $X$. The $\Sigma$-product is not separable since there are uncountably many factors in the product space $X$ and that each factor has at least two points. This idea had been implemented in this previous post.

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Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space $X$, let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For $F \in \mathcal{F}[X]$ and for any open subset $U$ of $X$, let $[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$. The sets $[F,U]$ over all $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space $X$ is uncountable, $\mathcal{F}[X]$ is not a separable space. We need to identify the $\mathcal{F}[X]$ that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space $X$ with a countable network, $\mathcal{F}[X]$ is CCC. Thus for any uncountable space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC space that is not separable. The following gives a few such examples.

$\mathcal{F}[\mathbb{R}]$

$\mathcal{F}[X]$ where $X$ is any uncountable, separable and metrizable space.

$\mathcal{F}[X]$ where $X$ is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space $X$ that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

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Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line $\mathbb{R}$. Let $X$ be any uncountable completely regular space. The product space $\mathbb{R}^X$ always has the CCC since it is a product of separable space. Now we single out a dense subspace of $\mathbb{R}^X$ for which there is a characterization for separability, namely the subspace $C(X)$, which is the set of all continuous functions from $X$ into $\mathbb{R}$. The subspace $C(X)$ as a topological space is usually denoted by $C_p(X)$. For a basic discussion of $C_p(X)$, see this previous post.

We know precisely when $C_p(X)$ is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space $C_p(X)$ is separable if and only if the domain space $X$ has a weaker (or coarser) separable metric topology (in other words, $X$ is submetrizable with a separable metric topology).

Based on the theorem, $C_p(X)$ is separable for any separable metric space $X$. Other examples of separable $C_p(X)$ include spaces $X$ that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus $C_p(X)$ would be separable if $X$ is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces $X$ that have no weaker separable metric topology.

• Any compact space $X$ that is not metrizable.
• The space $X=\omega_1$, the first uncountable ordinal with the order topology.
• Any space $X=C_p(Y)$ where $Y$ is not separable.

The function space $C_p(X)$ for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact $X$ are: the first uncountable successor ordinal $\omega_1+1$, the double arrow space, and the product space $\left\{0,1 \right\}^{\omega_1}$.

It can be shown that $C_p(\omega_1)$ is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space $C_p(Y)$ has a weaker (or coarser) separable metric topology if and only if the domain space $Y$ is separable.

Thus picking a non-separable space $Y$ would guarantee that $C_p(Y)$ has a weaker separable metric topology. As a result, $C_p(C_p(Y))$ is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

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Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space $X$ that is CCC and not separable from any one of the avenues discussed here, the function space $C_p(X)$ has no weaker separable metric topology (by Theorem 2) and the function space $C_p(C_p(X))$ is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

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Exercises

1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if $X$ is metrizable and CCC, then $X$ is separable.
2. Show that any dense subspace of a CCC space is also CCC.
3. Verify that the space $\Sigma(\left\{0,1 \right\}^{c})$ defined in Example 2 is dense in $X$ and is not separable.
4. Verify that the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ defined in Example 3 is CCC and not separable.
5. Verify that function space $C_p(\omega_1)$ mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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$\copyright \ 2016 \text{ by Dan Ma}$

# Cp(X) where X is a separable metric space

Let $\tau$ be an uncountable cardinal. Let $\prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau}$ be the Cartesian product of $\tau$ many copies of the real line. This product space is not normal since it contains $\prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1}$ as a closed subspace. However, there are dense subspaces of $\mathbb{R}^{\tau}$ are normal. For example, the $\Sigma$-product of $\tau$ copies of the real line is normal, i.e., the subspace of $\mathbb{R}^{\tau}$ consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of $\mathbb{R}^{\tau}$ that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space $C_p(X)$ where $X$ is a separable metrizable space.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

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$C_p(X)$ when $X$ is a separable metric space

In the remainder of the post, $X$ denotes a separable metrizable space. Then, $C_p(X)$ is more than normal. The function space $C_p(X)$ has the following properties:

• normal,
• Lindelof (hence paracompact and collectionwise normal),
• hereditarily Lindelof (hence hereditarily normal),
• hereditarily separable,
• perfectly normal.

All such properties stem from the fact that $C_p(X)$ has a countable network whenever $X$ is a separable metrizable space.

Let $L$ be a topological space. A collection $\mathcal{N}$ of subsets of $L$ is said to be a network for $L$ if for each $x \in L$ and for each open $O \subset L$ with $x \in O$, there exists some $A \in \mathcal{N}$ such that $x \in A \subset O$. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for $C_p(X)$, let $\mathcal{B}$ be a countable base for the domain space $X$. For each $B \subset \mathcal{B}$ and for any open interval $(a,b)$ in the real line with rational endpoints, consider the following set:

$[B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}$

There are only countably many sets of the form $[B,(a,b)]$. Let $\mathcal{N}$ be the collection of sets, each of which is the intersection of finitely many sets of the form $[B,(a,b)]$. Then $\mathcal{N}$ is a network for the function space $C_p(X)$. To see this, let $f \in O$ where $O=\bigcap_{x \in F} [x,O_x]$ is a basic open set in $C_p(X)$ where $F \subset X$ is finite and each $O_x$ is an open interval with rational endpoints. For each point $x \in F$, choose $B_x \in \mathcal{B}$ with $x \in B_x$ such that $f(B_x) \subset O_x$. Clearly $f \in \bigcap_{x \in F} \ [B_x,O_x]$. It follows that $\bigcap_{x \in F} \ [B_x,O_x] \subset O$.

Examples include $C_p(\mathbb{R})$, $C_p([0,1])$ and $C_p(\mathbb{R}^\omega)$. All three can be considered subspaces of the product space $\mathbb{R}^c$ where $c$ is the cardinality of the continuum. This is true for any separable metrizable $X$. Note that any separable metrizable $X$ can be embedded in the product space $\mathbb{R}^\omega$. The product space $\mathbb{R}^\omega$ has cardinality $c$. Thus the cardinality of any separable metrizable space $X$ is at most continuum. So $C_p(X)$ is the subspace of a product space of $\le$ continuum many copies of the real lines, hence can be regarded as a subspace of $\mathbb{R}^c$.

A space $L$ has countable extent if every closed and discrete subset of $L$ is countable. The $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ of the separable metric spaces $\left\{X_\alpha: \alpha \in A \right\}$ is a dense and normal subspace of the product space $\prod_{\alpha \in A} X_\alpha$. The normal space $\Sigma_{\alpha \in A} X_\alpha$ has countable extent (hence collectionwise normal). The examples of $C_p(X)$ discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Working with the function space Cp(X)

This post provides basic information about the space of real-valued continuous functions with the pointwise convergence topology. The goal is to discuss the setting and to define the standard basic open sets in the function space, providing background information for subsequent posts.

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Completely Regular Spaces

The starting point is a completely regular space. A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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Defining the Function Space $C_p(X)$

Let $X$ be a completely regular space. Let $C(X)$ be the set of all real-valued continuous functions defined on the space $X$. The set $C(X)$ is naturally a subspace of the product space $\prod_{x \in X} \mathbb{R}=\mathbb{R}^X$. Thus $C(X)$ can be endowed with the subspace topology inherited from the product space $\mathbb{R}^X$. When this is the case, the function space $C(X)$ is denoted by $C_p(X)$. The topology on $C_p(X)$ is called the pointwise convergence topology.

Now we need a good handle on the open sets in the function space $C_p(X)$. A basic open set in the product space $\mathbb{R}^X$ is of the form $\prod_{x \in X} U_x$ where each $U_x$ is an open subset of $\mathbb{R}$ such that $U_x = \mathbb{R}$ for all but finitely many $x \in X$ (equivalently $U_x \ne \mathbb{R}$ for only finitely many $x \in X$). Thus a basic open set in $C_p(X)$ is of the form:

$C(X) \cap \biggl(\prod_{x \in X} U_x \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where each $U_x$ is an open subset of $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$. In addition, when $U_x \ne \mathbb{R}$, we can take $U_x$ to be an open interval of the form $(a,b)$. To simplify notation, the basic open sets as described in (1) can also be notated by:

$\prod_{x \in X} U_x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1a)$

Thus when working with open sets in $C_p(X)$, we take $\prod_{x \in X} U_x$ to mean the set of all $f \in C(X)$ such that $f(x) \in U_x$ for each $x \in X$.

To make the basic open sets of $C_p(X)$ more explicit, (1) or (1a) is translated as follows:

$\bigcap_{x \in F} \ [x, O_x] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

where $F \subset X$ is a finite set, for each $x \in F$, $O_x$ is an open interval of $\mathbb{R}$, and $[x,O_x]$ is the set of all $f \in C(X)$ such that $f(x) \in O_x$.

There is another description of basic open sets that is useful. Let $f \in C_p(X)$. Let $F \subset X$ be finite. Let $\epsilon>0$. Let $B(f,F,\epsilon)$ be defined as follows:

$B(f,F,\epsilon)=\left\{g \in C(X): \forall \ x \in F, \lvert f(x)-g(x) \lvert< \epsilon \right\} \ \ \ \ \ \ \ (3)$

In proving results about $C_p(X)$, we can use basic open sets that are described in any one of the three forms (1), (2) and (3). If $U$ is a basic open subset of $C_p(X)$, as described in (1) or (1a), we use $supp(U)$ to denote the finite set of $x \in X$ such that $U_x \ne \mathbb{R}$. The set $supp(U)$ is called the support of $U$. The support for the basic open sets as described in (2) and (3) is already explicitly stated.

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Basic Discussion

The theory of $C_p(X)$ is a vast subject area. For a systematic introduction, see [1]. One fundamental theme in function space theory is the study of how properties of $X$ and $C_p(X)$ are related. The domain space $X$ and the function space $C(X)$ are not on the same footing. The domain $X$ only has a topological structure. The function space $C_p(X)$ carries a topology and two natural algebraic operations of addition and multiplication, making it a topological ring. In addition, $C_p(X)$ can be regarded as a topological group, or a linear topological space. In this post and in many subsequent posts, we narrow the focus to the topological properties of $X$ and $C_p(X)$, paying attention to the how the topological properties of $X$ and $C_p(X)$ are related.

In addition to the pointwise convergence topology, there are other topologies that can be defined on $C(X)$, e.g., the compact-open topology, the topology of uniform convergence and others. Both the pointwise convergence topology and the compact-open topology are examples of set-open topologies. In this post and in many of the subsequent posts, the focus is on the pointwise convergence topology, i.e., the subspace topology on $C(X)$ inherited from the product space.

The space $C_p(X)$ automatically inherits certain properties of the product space $\mathbb{R}^X$. Note that $C(X)$ is dense in $\mathbb{R}^X$. The product $\mathbb{R}^X$ has the countable chain condition (CCC) since it is a product of separable spaces. Hence $C_p(X)$ always has the CCC, i.e., there are no uncountably many pairwise disjoint open subsets of $C_p(X)$, regardless what the domain space $X$ is. One consequence of the CCC is that $C_p(X)$ is paracompact if and only if $C_p(X)$ is Lindelof.

It is well known that $\mathbb{R}^X$ is separable if and only if the cardinality of $X$ $\le$ continuum. Since $C(X)$ is dense in $\mathbb{R}^X$, $C_p(X)$ is not separable if the cardinality of $X$ $>$ continuum. Thus $C_p(X)$ is one way to get a CCC space that is not separable. There are non-separable $C_p(X)$ where the cardinality of $X$ $\le$ continuum. Obtaining such $C_p(X)$ would require more than the properties of the product space $\mathbb{R}^X$; using properties of $X$ would be necessary.

The properties of $C_p(X)$ discussed so far are inherited from the product space. Refer to chapter one of [1] for other elementary properties of $C_p(X)$. See this post for a discussion of $C_p(X)$ where $X$ is a separable metric space. See this post about a consequence of normality of $C_p(X)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A factorization theorem for products of separable spaces

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $f:X \rightarrow T$ be a continuous function where $T$ is a topological space. In this post, we discuss what it means for the continuous function $f$ to depend on countably many coordinates and then discuss some conditions that we can impose on the product space and on the range space $T$ to ensure that every continuous $f$ defined on the product space will depend on countably many coordinates. This notion of a continuous function depending on countably many coordinates is equivalent to factoring the continuous function into the composition of a projection map and a continuous function defined on a countable subproduct (see Lemma 1 below).

Let’s set some notation about the product space we work with in this post. Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. For any $B \subset A$, $\pi_B$ is the natural projection from the full product space $X=\prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in B} X_\alpha$. Standard basic open sets of $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is open in $X_\alpha$ and that $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the finite set of $\alpha \in A$ where $O_\alpha \ne X_\alpha$.

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Factoring a Continuous Map

The function $f$ is said to depend on countably many coordinates if there exists a countable set $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$. Suppose $f$ is instead defined on a subspace $Y$ of $X$. The function $f$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.

We have the following lemmas.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $f:X \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \prod_{\alpha \in B} X_\alpha \rightarrow T$ is continuous.

Lemma 1a

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space. Let $T$ be a topological space. Let $Y$ be a dense subspace of $X$. Let $f:Y \rightarrow T$ be continuous. Then the following are equivalent.

1. There exists a countable $B \subset A$ such that for any $x,y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $f(x)=f(y)$.
2. There exists a countable $B \subset A$ such that $f=g \circ \pi_B$ where $g: \pi_B(Y) \rightarrow T$ is continuous.

It is straightforward to verify Lemma 1 and Lemma 1a. We use condition 1 to define what it means for a function to be dependent on countably many coordinates. Both lemmas indicate that either condition is a valid definition. These two lemmas also indicate why the notion being discussed can be called a factorization notion.

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When a Continuous Map Can Be Factored

We discuss some conditions that we can place on the product space $X=\prod_{\alpha \in A} X_\alpha$ and on the range space $T$ so that any continuous map depends on countably many coordinates. We prove the following theorem.

Theorem 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space (i.e. having a countable base). Then for any dense subspace $Y$ of $X$, any continuous function $f:Y \rightarrow T$ depends on countably many coordinates, i.e., either one of the conditions in Lemma 1a holds.

Before stating the main theorem, we need one more lemma. Let $W \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$.

When we try to determine whether a function $f:Y \rightarrow T$, where $Y \subset X$, can be factored, we will need to decide whether a set $W \subset Y$ depends on countably many coordinates. Let $W \subset Y \subset X=\prod_{\alpha \in A} X_\alpha$. The set $W$ is said to depend on countably many coordinates if there exists a countable $B \subset A$ such that for any $x \in W$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in W$. We have the following lemma.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space with the countable chain condition. Let $Y$ be a dense subspace of $X$.

1. Let $U$ be an open subset of $X$. Then $\overline{U}$ depends on countably many coordinates.
2. Let $W$ be an open subset of $Y$. Then $\overline{W}$ depends on countably many coordinates (closure in $Y$).

Proof of Lemma 2
Proof of Part 1
Let $U \subset X$ be open. Let $\mathcal{B}$ be a collection of pairwise disjoint open subsets of the open set $U \subset X$ such that $\mathcal{B}$ is maximal with this property, i.e., if you throw one more open set into $\mathcal{B}$, it will be no longer pairwise disjoint. Let $V=\bigcup \mathcal{B}$. Since $\mathcal{B}$ is maximal, $\overline{V}=\overline{U}$. Since $X$ has the countable chain condition, $\mathcal{B}$ is countable.

Let $B=\bigcup \left\{supp(O): O \in \mathcal{B} \right\}$. The set $B$ is a countable subset of $A$ since $B$ is the union of countably many finite sets. We have the following claims.

Claim 1
The open set $V$ depends on the coordinates in $B$.

Let $x \in V$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $y \in V$. Firstly, $x \in O$ for some $O \in \mathcal{B}$. It follows that $x_\alpha=y_\alpha$ for all $\alpha \in supp(O)$. Thus $y \in O \subset V$. This completes the proof of Claim 1.

Claim 2
The set $\overline{V}$ depends on the coordinates in $B$.

Let $x \in \overline{V}$ and $y \in X$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show $y \in \overline{V}$. To this end, let $O=\prod_{\alpha \in A} O_\alpha$ be a standard basic open set with $y \in O$. The goal is to find some $q \in O \cap V$. Define $G=\prod_{\alpha \in A} G_\alpha$ such that $G_\alpha=O_\alpha$ for all $\alpha \in B$ and $G_\alpha=X_\alpha$ for all $\alpha \in A-B$. Then $x \in G$. Since $x \in \overline{V}$, there exists some $p \in V \cap G$. Define $q$ such that $q_\alpha=p_\alpha$ for all $\alpha \in B$ and $q_\alpha=y_\alpha$ for all $\alpha \in A-B$. Since $supp(V) \subset B$, $q \in V$. On the other hand, $q \in O$. This completes the proof of Claim 2.

As noted above, $\overline{V}=\overline{U}$. Thus $\overline{U}$ depends on countably many coordinates, namely the coordinates in the set $B$. This completes the proof of Part 1.

Proof of Part 2
For any $S \subset X$, let $\overline{S}$ denote the closure of $S$ in $Y$. Let $Cl_X(S)$ denote the closure of $S$ in $X$. Let $W \subset Y$ be open. Let $W_1$ be open in $X$ such that $W=W_1 \cap Y$. By Part 1, $Cl_X(W_1)$ depends on countably many coordinates, say the coordinates in the countable set $B \subset A$. This means that for any $x \in Cl_X(W_1)$ and for any $y \in X$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. Thus for any $x \in \overline{W}$ and for any $y \in Y$, if $x_\alpha=y_\alpha$ for all $\alpha \in B$, then $y \in Cl_X(W_1)$. If we have $y \in \overline{W}$, then we are done. So we only need to show that if $y \in Y$ and $y \in Cl_X(W_1)$, then $y \in \overline{W}$. This is why we need to assume $Y$ is dense in $X$.

Let $y \in Y$ and $y \in Cl_X(W_1)$. Let $C$ be an open subset of $Y$ with $y \in C$. There exists an open subset $D$ of $X$ such that $C=D \cap Y$. Then $D \cap Cl_X(W_1) \ne \varnothing$. Note that $D \cap W_1$ is open and $D \cap W_1 \ne \varnothing$. Since $Y$ is dense in $X$, $D \cap W_1$ must contain points of $Y$. These points of $Y$ are also points of $W$. Thus $C$ contains points of $W$. It follows that $y \in \overline{W}$. This concludes the proof of Part 2. $\blacksquare$

Proof of Theorem 1
Let $Y$ be a dense subspace of $X=\prod_{\alpha \in A} X_\alpha$. Let $f:Y \rightarrow T$ be continuous. Let $\mathcal{M}$ be a countable base for the separable metrizable space $T$. By Lemma 2 Part 2, for each $M \in \mathcal{M}$, $\overline{f^{-1}(M)}$ depends on countably many coordinates, say the countable set $B_M$. Let $B=\bigcup_{M \in \mathcal{M}} B_M$.

We claim that $B$ is a countable set of coordinates we need. Let $x,y \in Y$ such that $x_\alpha=y_\alpha$ for all $\alpha \in B$. We need to show that $f(x)=f(y)$. Suppose $f(x) \ne f(y)$. Choose $\left\{M_1,M_2,M_3,\cdots \right\} \subset \mathcal{M}$ such that

• $\left\{f(x) \right\}=\bigcap_{j=1}^\infty M_j=\bigcap_{j=1}^\infty \overline{M_j}$
• $\overline{M_{j+1}} \subset M_j$ for each $j$

This is possible since $T$ is a second countable space. Then $f(y) \notin \overline{M_{k}}$ for some $k$. Furthermore, $y \notin f^{-1}(\overline{M_{k}})$. Since $f$ is continuous, $\overline{f^{-1}(M_{k})} \subset f^{-1}(\overline{M_{k}})$. Therefore, $y \notin \overline{f^{-1}(M_{k})}$. On the other hand, $\overline{f^{-1}(M_{k})}$ depends on the countably many coordinates in $B_{M_k}$. We assume above that $x_\alpha=y_\alpha$ for all $\alpha \in B$. Thus $x_\alpha=y_\alpha$ for all $\alpha \in B_{M_k}$. This means that $y \in \overline{f^{-1}(M_{k})}$, a contradiction. It must be that case that $f(x)=f(y)$. $\blacksquare$

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Another Version

We state another version of Theorem 1 that will be useful in some situations.

Theorem 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product space such that each factor $X_\alpha$ is a separable space. Let $T$ be a second countable space. Let $Y$ be a dense subspace of $X$. Let $f:Y \times Y \rightarrow T$ be any continuous function. Then the function $f$ depends on countably many coordinates, which means either one of the following two conditions:

1. There exists a countable set $C \subset A$ such that for any $(x,y),(p,q) \in Y \times Y$, if $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$, then $f(x,y)=f(p,q)$.
2. There exists a countable set $C \subset A$ and there exists a continuous $g:\pi_C(Y) \times \pi_C(Y) \rightarrow T$ such that $f=g \circ (\pi_C \times \pi_C)$.

The map $\pi_C \times \pi_C$ is the projection map from $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ into the subproduct $\prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in B} X_\alpha$ defined by $(\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y))$. In Theorem 2, we only need to consider $\pi_C \times \pi_C$ being defined on the subspace $Y \times Y$.

Theorem 2 follows from Theorem 1. It is only a matter of fitting Theorem 2 in the framework of Theorem 1. Note that the product $\prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$ is identical to the product $\prod_{\alpha \in A \cup A^*} X_\alpha$ where $A^*$ is a disjoint copy of the index set $A$. For $(x,y) \in \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha$, let $x \times y \in \prod_{\alpha \in A \cup A^*} X_\alpha$ be defined by $(x \times y)_\alpha=x_\alpha$ for all $\alpha \in A$ and $(x \times y)_\alpha=y_\alpha$ for all $\alpha \in A^*$.

With the identification of $(x,y)$ with $x \times y$, we have a setting that fits Theorem 1. The product $\prod_{\alpha \in A \cup A^*} X_\alpha$ is also a product of separable spaces. The set $Y \times Y$ is a dense subspace of the product $\prod_{\alpha \in A \cup A^*} X_\alpha$. In this new setting, we view a point in $Y \times Y$ as $x \times y$. The map $f:Y \times Y \rightarrow T$ is still a continuous map. We can now apply Theorem 1.

Let $B \subset A \cup A^*$ be a countable set such that for all $x \times y,p \times q \in Y \times Y$, if $(x \times y)_\alpha=(p \times q)_\alpha$ for all $\alpha \in B$, then $f(x \times y)=f(p \times q)$. Specifically, if $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$, then $f(x,y)=f(p,q)$.

Choose a countable set $C \subset A$ such that $B \cap A \subset C$ and $B \cap A^* \subset C^*$. Here, $C^*$ is the copy of $C$ in $A^*$. We claim that $C$ is a countable set we need in condition 1 of Theorem 2. Let $(x,y),(p,q) \in Y \times Y$ such that $x_\alpha=p_\alpha$ and $y_\alpha=q_\alpha$ for all $\alpha \in C$. This implies that $x_\alpha=p_\alpha$ for all $\alpha \in B \cap A$ and $y_\alpha=q_\alpha$ for all $\alpha \in B \cap A^*$. Then $f(x,y)=f(p,q)$. Thus condition 1 of Theorem 2 holds. It is also straightforward to verify that Condition 1 and Condition 2 are equivalent.

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Remarks

The notion of factorizing a continuous map defined on a product space is an old topic. Theorem 1 discussed in this post is based on Theorem 4 found in [6]. Theorem 4 found in [6] is to factor continuous maps defined on a product of separable spaces. Theorem 1 in this post is modified to consider continuous maps defined on a dense subspace of a product of separable spaces. This modification will make it more useful. The references listed below represent a small sample of papers or books that have involves theorems of factoring functions defined on products. The work in [3] and [5] have more systematic treatment.

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Reference

1. Brandenburg H., Husek M., On mappings from products into developable spaces, Topology Appl., 26, 229-238, 1987.
2. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Engelking R., On functions defined on Cartesian products, Fund. Math., 59, 221-231, 1966.
4. Keesling J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble N., Ulmer M., Factoring functions on Cartesian products, Trans. Amer. Math. Soc., 163, 329-339, 1972.
6. Ross K. A., Stone A. H., Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A lemma dealing with normality in products of separable metric spaces

In this post we prove a lemma that is a great tool for working with product spaces of separable metrizable spaces. As an application of the lemma, we give an alternative proof for showing the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$ where each $X_\alpha$ is a separable and metrizable space. The lemma we discuss here is a tool that can shed some light on normality of dense subspaces of the product space $X$. The lemma is stated in two equivalent forms (Lemma 1 and Lemma 2).

Before stating the lemmas, let’s fix some notations. For any $B \subset A$, the map $\pi_B$ is the natural projection from the full product $X=\prod_{\alpha \in A} X_\alpha$ to the subproduct $\prod_{\alpha \in B} X_\alpha$. The standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$ are of the form $\prod_{\alpha \in A} O_\alpha$ where $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. We use $supp(\prod_{\alpha \in A} O_\alpha)$ to denote the set of finitely many $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

Given a space $W$, and given $F,G \subset W$, the sets $F$ and $G$ are separated if $F \cap \overline{G}=\varnothing=\overline{F} \cap G$.

Lemma 1

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. For any sets $H,K \subset Y$, the following two conditions are equivalent:

1. There exist disjoint open subsets $U$ and $V$ of $Y$ such that $H \subset U$ and $K \subset V$.
2. There exists a countable $B \subset A$ such that the sets $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$.

Lemma 2

Let $X=\prod_{\alpha \in A} X_\alpha$ be a product of separable metrizable spaces. Let $Y$ be a dense subspace of $X$. Then $Y$ is normal if and only if for each pair of disjoint closed subsets $H$ and $K$ of $Y$, there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$.

If Lemma 1 holds, it is clear that Lemma 2 holds. We prove Lemma 1. The lemmas indicate that to separate disjoint sets in the full product, it suffices to separate in a countable subproduct. In this sense normality in dense subspaces of a product of separable metrizable spaces only depends on countably many coordinates.

This lemma seems to have been around for a long time. We cannot find any reference of this lemma in Engelking’s topology textbook (see [4]). We found three references. One is Corson’s paper (see [3]), in which the lemma is mentioned in relation to the non-normality of $\mathbb{N}^{\omega_1}$ and is attributed to a paper of M. Bockstein in 1948. Another is a paper of Baturov (see [2]), in which the lemma is used to prove a theorem about normality in dense subspace of $M^{\omega_1}$ where $M$ is a separable metric space. In [2] the lemma is attributed to Uspenskii. Another reference is Arkhangelskii’s book on function space (see Lemma I.6.1 on p. 43 in [1]) where the lemma is used in proving some facts about normality in function spaces $C_p(X)$.

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Proof of Lemma 1

$1 \Longrightarrow 2$
Let $U$ and $V$ be disjoint open subsets of $Y$ with $H \subset U$ and $K \subset V$. Let $U_1$ and $V_1$ be open subsets of $X$ such that $U=U_1 \cap Y$ and $V=V_1 \cap Y$. Since $Y$ is dense in $X$, $U_1 \cap V_1=\varnothing$.

Let $\mathcal{U}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $U_1$. Let $\mathcal{V}$ be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of $V_1$. These two collections can be obtained using a Zorn lemma argument. The product space $X$ has the countable chain condition since it is a product of separable spaces. So both $\mathcal{U}$ and $\mathcal{V}$ are countable. Let $B$ be the union of finite sets each one of which is a $supp(O)$ where $O \in \mathcal{U} \cup \mathcal{V}$. The set $B$ is countable too.

Let $U^*=\cup \mathcal{U}$ and $V^*=\cup \mathcal{V}$. Note that $U^* \cap V^*=\varnothing$. We have the following observations:

$\pi^{-1}_B(\pi_B(U^*))=U^* \subset U_1$ and $\pi^{-1}_B(\pi_B(V^*))=V^* \subset V_1$

The above observations lead to the following observations:

$\pi^{-1}_B(\pi_B(U^*)) \cap \pi^{-1}_B(\pi_B(V^*)) \subset U_1 \cap V_1=\varnothing$

implying that $\pi_B(U^*) \cap \pi_B(V^*)=\varnothing$. Both $\pi_B(U^*)$ and $\pi_B(V^*)$ are open subsets of $\pi_B(X)$ and are dense in $\pi_B(X)$, respectively.

We claim that $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$. Suppose that $y \in \pi_B(U_1) \cap \pi_B(V_1)$. Then $\pi_B(V_1)$ contains a point of $\pi_B(U^*)$, say $t$. With $t \in \pi_B(U^*)$, $t=\pi_B(q)$ for some $q \in O$ where $O \in \mathcal{U}$. Note that $supp(O) \subset B$. Thus $\pi^{-1}_B(\pi_B(q))=\pi^{-1}_B(t)=O \subset U_1$. On the other hand, $t \in \pi_B(V_1)$ implies that $t=\pi_B(w)$ for some $w \in V_1$. It follows that $w \in U_1 \cap V_1$, a contradiction. Therefore $\pi_B(U_1) \cap \pi_B(V_1)=\varnothing$.

We have $\pi_B(H) \subset \pi_B(U) \subset \pi_B(U_1)$ and $\pi_B(K) \subset \pi_B(V) \subset \pi_B(V_1)$. This implies that $\overline{\pi_B(H)} \cap \pi_B(K)=\varnothing$ and $\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$ (closure in $\pi_B(X)$). Then $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$ as well. This concludes the proof for the $1 \Longrightarrow 2$ direction.

$2 \Longrightarrow 1$
Suppose that $B \subset A$ is countable such that $\pi_B(H)$ and $\pi_B(K)$ are separated in the space $\pi_B(Y)$. Note that $\pi_B(H) \subset \pi_B(Y)$ and $\pi_B(K) \subset \pi_B(Y)$. Then we have the following:

$\pi_B(H) \cup \pi_B(K) \subset \pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}) \ \ \ \ \text{closures in } \pi_B(Y)$

Consider $W=\pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)})$. The space $W$ is an open subspace of $\pi_B(Y)$. Furthermore, $\pi_B(Y)$ is a subspace of $\prod_{\alpha \in B} X_\alpha$, which is a separable and metrizable space. Thus the space $W$ is metrizable and hence normal.

For $L \subset W$, let $Cl_W(L)$ denote the closure of $L$ in the space $W$. Note that $Cl_W(\pi_B(H))$ and $Cl_W(\pi_B(K))$ are disjoint and closed sets in $W$. Let $G_H$ and $G_K$ be disjoint open subsets of $W$ such that $Cl_W(\pi_B(H)) \subset G_H$ and $Cl_W(\pi_B(K)) \subset G_K$. Then $\pi^{-1}_B(G_H) \cap Y$ and $\pi^{-1}_B(G_K) \cap Y$ are disjoint open subsets of $Y$ such that $H \subset \pi^{-1}_B(G_H) \cap Y$ and $K \subset \pi^{-1}_B(G_H) \cap Y$. $\blacksquare$

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Remark

The proof of Lemma 1 does not need the full strength of separable metric in each factor of the product space. The above proof only makes two assumptions about the product space: the product space $X=\prod_{\alpha \in A} X_\alpha$ has the countable chain condition (CCC) and that any countable subproduct is normal, i.e., $\prod_{\alpha \in B} X_\alpha$ is normal for any countable $B \subset A$.

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Example

As an application of the above lemma, we give another proof of the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers. See this post for a version of A. H. Stone’s original proof.

Let $\mathbb{N}$ be the set of all nonnegative integers and let $\omega_1$ be the first uncountable ordinal (i.e. the set of all countable ordinals). We provide an alternative proof that $\mathbb{N}^{\omega_1}$ is not normal. In A. H. Stone’s proof, the following disjoint closed sets cannot be separated in $\mathbb{N}^{\omega_1}$:

$H=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 0, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

$K=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 1, x_\alpha=n \text{ for at most one } \alpha<\omega_1 \right\}$

We can also use Lemma 1 to show that $H$ and $K$ cannot be separated. Note that for each countable $B \subset \omega_1$, the sets $\pi_B(H)$ and $\pi_B(K)$ have non-empty intersection. Hence they cannot be separated in $\pi_B(\mathbb{N}^{\omega_1})$. By Lemma 1, $H$ and $K$ cannot be separated in the full product space $\mathbb{N}^{\omega_1}$.

To see that $\pi_B(H) \cap \pi_B(K) \ne \varnothing$, choose a function $g:\omega_1 \rightarrow \mathbb{N}$ such that $g^{-1}(0) \cap B=\varnothing$. Let $g_B:B \rightarrow \mathbb{N}$ be defined by $g_B(\alpha)=g(\alpha)$ for all $\alpha \in B$. Then $g_B \in \pi_B(H) \cap \pi_B(K)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Looking for a closed and discrete subspace of a product space

I had long suspected that there probably is an uncountable closed and discrete subset of the product space of uncountably many copies of the real line. Then I found the statement in the Encyclopedia of General Topology (page 76 in [3]) that “for every infinite cardinal $\mathcal{K}$, the product $D(\mathcal{K})^{2^{\mathcal{K}}}$ includes $D(2^{\mathcal{K}})$ as a closed subspace” where $D(\tau)$ is the discrete space of cardinality of $\tau$. When $\mathcal{K}=\aleph_0$ (the first infinite cardinal), there is a closed and discrete subset of cardinality $c=2^{\aleph_0}$ in the product space $\mathbb{N}^{c}$ (the product space of continuum many copies of a countable discrete space). Despite the fact that this product space is a separable space, a closed and discrete set of cardinality continuum is hiding in the product space $\mathbb{N}^{c}$. What is more amazing is that this result gives us a glimpse into the working of the product topology with uncountably many factors. There are easily defined discrete subspaces of $\mathbb{N}^{c}$. But these discrete subspaces are not closed in the product space, making the result indicated here a remarkable one.

The Encyclopedia of General Topology points to two references [2] and [4]. I could not find these papers online. It turns out that Engelking, the author of [2], included this fact as an exercise in his general topology textbook (see Exercise 3.1.H (a) in [1]). This post presents a proof of this fact based on the hints that are given in [1]. To make the argument easier to follow, the proof uses some of the hints in a slightly different form.

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The Exercise

Let $I=[0,1]$ be the closed unit interval. Let $X=I$ be the unit interval with the discrete topology. Let $\omega$ be the set of all nonnegative integers with the discrete topology. Let $Y=\prod_{t \in I} Y_t$ where each $Y_t=\omega$. We can also denote $Y$ by $\omega^I$. The problem is to show that the discrete space $X$ can be embedded as a closed and discrete subspace of $Y$.

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A Solution

For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that

• $t \in O_{t,j}$ for each $j$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$),
• $\left\{t \right\}=\bigcap \limits_{j=1}^\infty O_{t,j}$.

For any $t \in I-\left\{0,1 \right\}$, we can make $O_{t,j}$ open intervals of the form $(a,b)$. For $t=0$, $O_{0,j}$ have the form $[0,b)$. For $t=1$, $O_{1,j}$ have the form $(a,1]$.

The above sequences of open intervals help define a homeomorphic embedding of the discrete space $X$ into the product $Y$. For each $t \in I$, define the function $f_t:I \rightarrow \omega$ by letting:

$f_t(x) = \begin{cases} 0 & \mbox{if } x=t \\ 1 & \mbox{if } x \in I-O_{t,1} \\ 2 & \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \cdots \\ j & \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \text{etc} \end{cases}$

for each $x \in X$. We now define the embedding $E:X \rightarrow Y=\omega^I$ by letting:

$E(x)=< f_t(x) >_{t \in I}$

for each $x \in X$. For each point $x \in X$, $E(x)$ is the point in the product space such that the $t$ coordinate of $E(x)$ is the value of the function $f_t$ evaluated at $x$. Let $\mathcal{F}=\left\{f_t: t \in I \right\}$.

The mapping $E$ is an evaluation map (called diagonal map in [1]). It is a homeomorphism if the following three conditions are met:

• If each $f_t \in \mathcal{F}$ is continuous, then $E$ is continuous.
• If the family of functions $\mathcal{F}$ separates points in $X$, then $E$ is injective (i.e. a one-to-one function).
• If $\mathcal{F}$ separates points from closed sets in $X$, then the inverse $E^{-1}$ is also continuous.

The first point is easily seen. Note that both the domain and the range of $f_t:X \rightarrow \omega$ have the discrete topology. Thus these functions are continuous. To see the second point, let $p,q \in X$ with $p \ne q$. Note that $f_p(p)=0$ while $f_p(q) \ne 0$.

Since $X$ is discrete, any subset of $X$ is closed. To see the third point, let $C \subset X$ and $p \notin C$. Once again, $f_p(p)=0$ while $f_p(x) \ne 0$ for all $x \in C$. Clearly $f_p(p) \notin \overline{f(C)}$ (closure in the discrete space $\omega$). Thus $E^{-1}$ is also continuous. For more details about why $E$ is an embedding, see the previous post called The Evaluation Map.

Let $W=E(X)$. Since $E$ is a homeomorphism, $W$ is a discrete subspace of the product space $Y$. We only need to show $W$ is closed in $Y$. Let $k \in Y$ such that $k=< k_t >_{t \in I} \ \notin W$. We show that there is an open neighborhood of $k$ that misses the set $W$. There are two cases to consider. One is that $k_r \ne 0$ for all $r \in I$. The other is that $k_r=0$ for some $r \in I$. The first case is more involved.

Case 1
Suppose that $k_r \ne 0$ for all $r \in I$. First define a local base $\mathcal{B}_{k}$ of the point $k$. Let $H \subset I$ be finite and let $G_{H}$ be defined by:

$G_{H}=\left\{b \in Y: \forall \ h \in H, b_h=k_h \right\}$

Let $\mathcal{B}_{k}$ be the set of all possible $G_{H}$. For an arbitrary $G_{H}$, consider $E^{-1}(G_H)$. By definition, $E^{-1}(G_H)=\left\{c \in I: E(c) \in G_H \right\}$. To make the argument below easier to see, let’s further describe $E^{-1}(G_H)$.

\displaystyle \begin{aligned} E^{-1}(G_H)&=\left\{c \in I: E(c) \in G_H \right\} \\&=\left\{c \in I: \forall \ h \in H, f_h(c)=k_h \ne 0 \right\} \\&=\left\{c \in I: \forall \ h \in H, c \in I-O_{h,k_h} \right\} \\&=\bigcap \limits_{h \in H} I-O_{h,k_h} \\&=I-\bigcup \limits_{h \in H} O_{h,k_h} \end{aligned}

The last description above indicates that if $x \in X$ and if $x \in E^{-1}(G_H)$, then $x \notin \bigcup \limits_{h \in H} O_{h,k_h}$. To wrap up Case 1, we would like to produce one particular $H$. Consider the open cover $\left\{O_{t,k_t}: t \in I \right\}$ of $I$. Since $I$ is compact in the usual topology, there is a finite $H \subset I$ such that $I=\bigcup \limits_{h \in H} O_{h,k_h}$. This means that for this particular finite set $H$, $E^{-1}(G_H)=\varnothing$. Putting it in another way, the open neighborhood $B_H$ of $k$ contains no point of $E(x)$. The proof for Case 1 is completed.

Case 2
Suppose that $k_r=0$ for some $r \in I$. Since $k \notin W=E(X)$, in particular $k \ne E(r)=< f_t(r) >_{t \in I}$. So for some $q \in I$, $k_q \ne f_q(r)$. Now define the following open set containing $k$.

$G=\left\{b \in Y: b_r=0 \text{ and } b_q=k_q \right\}$

Note that $E(r) \notin G$ since $k_q \ne f_q(r)$. Furthermore, for each $p \in I-\left\{r \right\}$, $E(p) \notin G$ since $f_r(p) \ne 0$. Thus $G$ is an open neighborhood of $k$ containing no point of $E(X)$. The proof for Case 2 is completed.

We have shown that the image of the discrete space $X$ under the homeomorphism $E$ is closed in the product space $Y=\prod_{t \in I} Y_t=\omega^I$.

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Comments

The preceding proof shows that the product space of continuum many copies of $\omega$ contains a closed and discrete subspace of cardinality continuum. This is a remarkable result. At a glance, it is not entirely clear that a closed and discrete set of this large size can be found in the product space in question.

One immediate consequence is that the product space $Y=\prod_{t \in I} Y_t=\omega^I$ is not normal since it is a separable space. By Jones’ lemma, any separable normal space cannot have a closed and discrete subset of cardinality continuum. However, if the goal is only to show non-normality, we only need to show that $\omega^{\omega_1}$ is not normal (a proof is found in this post). Thus the value of the preceding proof is to demonstrate how to produce a closed and discrete subspace of cardinality continuum in the product space in question.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., On the double circumference of Alexandroff, Bull. Acad. Polon. Sci., 16, 629-634, 1968.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Juhasz, I., On closed discrete subspace of product spaces, Bull. Acad. Polon. Sci., 17, 219-223, 1969.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A theorem about CCC spaces

It is a well known result in general topology that in any regular space with the countable chain condition, paracompactness and the Lindelof property are equivalent. The proof of this result hinges on one theorem about the spaces with the countable chain condition. In this post we are to put the spotlight on this theorem (Theorem 1 below) and then use it to prove a few results. These results indicate that in a space with the countable chain condition with some weaker covering property is either Lindelof or paracompact.

This post is centered on a theorem about the CCC property (Theorem 1 and Theorem 1a below). So it can be considered as a continuation of a previous post on CCC called Some basic properties of spaces with countable chain condition. The results that are derived from Theorem 1 are also found in [2]. But the theorem concerning CCC is only a small part of that paper among several other focuses. In this post, the exposition is to explain several interesting theorems that are derived from Theorem 1. One of the theorems is the statement that every locally compact metacompact perfectly normal space is paracompact, a theorem originally proved by Arhangelskii (see Theorem 11 below).

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CCC Spaces

All spaces under consideration are at least $T_1$ and regular. A space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if $X$ has the CCC, we also say that $X$ is a CCC space or $X$ is CCC.

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A Theorem about CCC Spaces

The theorem of CCC spaces we want to discuss has to do with collections of open sets that are “nice”. We first define what we mean by nice. Let $\mathcal{A}$ be a collection of non-empty subsets of the space $X$. The collection $\mathcal{A}$ is said to be point-finite (point-countable) if each point of $X$ belongs to only finitely (countably) many sets in $\mathcal{A}$.

Now we define what we mean by “nice” collection of open sets. The collection $\mathcal{A}$ is said to be locally finite (locally countable) at a point $x \in X$ if there exists an open set $O \subset X$ with $x \in O$ such that $O$ meets at most finitely (countably) many sets in $\mathcal{A}$. The collection $\mathcal{A}$ is said to be locally finite (locally countable) if it is locally finite (locally countable) at each $x \in X$.

The property of being a separable space implies the CCC. The reverse is not true. However the CCC property is still a very strong property. The CCC property is equivalent to the property that if a collection of non-empty open sets is “nice” on a dense set of points, then the collection of open sets is a countable collection. The following is a precise statement.

Theorem 1

Let $X$ be a CCC space. Then if $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The collections of open sets in the above theorem do not have to be open covers. However, if they are open covers, the theorem can tie CCC spaces with some covering properties. As long as the space has the CCC, any open cover that is locally-countable on a dense set must be countable. Looking at it in the contrapositive angle, in a CCC space, any uncountable open cover is not locally-countable in some open set.

Proof of Theorem 1
Let $\mathcal{U}$ be a collection of open subsets of $X$ such that the set $D(\mathcal{U})$ as defined above is dense in the open subspace $\bigcup \mathcal{U}$. We show that $\mathcal{U}$ is countable. Suppose not.

For each $U \in \mathcal{U}$, since $U \cap D(\mathcal{U}) \ne \varnothing$, we can choose a non-empty open set $f(U) \subset U$ such that $f(U)$ has non-empty intersection with only countably many sets in $\mathcal{U}$. Let $\mathcal{U}_f$ be the following collection:

$\mathcal{U}_f=\left\{f(U): U \in \mathcal{U} \right\}$

For $H,K \in \mathcal{U}_f$, by a chain from $H$ to $K$, we mean a finite collection

$\left\{W_1,W_2,\cdots,W_n \right\} \subset \mathcal{U}_f$

such that $H=W_1$, $K=W_n$ and $W_j \cap W_{j+1} \ne \varnothing$ for any $1 \le j . For each open set $W \in \mathcal{U}_f$, define $\mathcal{C}(W)$ and $\mathcal{E}(W)$ as follows:

$\mathcal{C}(W)=\left\{V \in \mathcal{U}_f: \text{there exists a chain from } W \text{ to } V \right\}$

$\mathcal{E}(W)=\bigcup \mathcal{C}(W)$

One observation we make is that for $W_1,W_2 \in \mathcal{U}_f$, if $\mathcal{E}(W_1) \cap \mathcal{E}(W_2) \ne \varnothing$, then $\mathcal{C}(W_1)=\mathcal{C}(W_2)$ and $\mathcal{E}(W_1)=\mathcal{E}(W_2)$. So the distinct $\mathcal{E}(W)$ are pairwise disjoint. Because the space $X$ has the CCC, there can be only countably many distinct open sets $\mathcal{E}(W)$. Thus there can be only countably many distinct collections $\mathcal{C}(W)$.

Note that each $\mathcal{C}(W)$ is a countable collection of open sets. Each $V \in \mathcal{U}_f$ meets only countably many open sets in $\mathcal{U}$. So each $V \in \mathcal{U}_f$ can meet only countably many sets in $\mathcal{U}_f$, since for each $V \in \mathcal{U}_f$, $V \subset U$ for some $U \in \mathcal{U}$. Thus for each $W \in \mathcal{U}_f$, in considering all finite-length chain starting from $W$, there can be only countably many open sets in $\mathcal{U}_f$ that can be linked to $W$. Thus $\mathcal{C}(W)$ must be countable. In taking the union of all $\mathcal{C}(W)$, we get back the collection $\mathcal{U}_f$. Thus we have:

$\mathcal{U}_f=\bigcup \limits_{W \in \mathcal{U}_f} \mathcal{C}(W)$

Because the space $X$ is CCC, there are only countably many distinct collections $\mathcal{C}(W)$ in the above union. Each $\mathcal{C}(W)$ is countable. So $\mathcal{U}_f$ is a countable collection of open sets.

Furthermore, each $U \in \mathcal{U}$ contains at least one set in $\mathcal{U}_f$. From the way we choose sets in $\mathcal{U}_f$, we see that for each $V \in \mathcal{U}_f$, $V=f(U) \subset U$ for at most countably many $U \in \mathcal{U}$. The argument indicates that we have a one-to-countable mapping from $\mathcal{U}_f$ to $\mathcal{U}$. Thus the original collection $\mathcal{U}$ must be countable. $\blacksquare$

The property in Theorem 1 is actually equivalent to the CCC property. Just that the proof of Theorem 1 represents the hard direction that needs to be proved. Theorem 1 can be expanded to be the following theorem.

Theorem 1a

Let $X$ be a space. Then the following conditions are equivalent.

1. The space $X$ has the CCC.
2. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

3. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that $\mathcal{U}$ is locally-countable at every point in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The direction $1 \rightarrow 2$ has been proved above. The directions $2 \rightarrow 3$ and $3 \rightarrow 1$ are straightforward.

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Tying Theorem 1 to “Nice” Open Covers

One easy application of Theorem 1 is to tie it to locally-finite and locally-countable open covers. We have the following theorem.

Theorem 2

In any CCC space, any locally-countable open cover must be countable. Thus any locally-finite open cover must also be countable.

Theorem 2 gives the well known result that any CCC paracompact space is Lindelof (see Theorem 5 below). In fact, Theorem 2 gives the result that any CCC para-Lindelof space is Lindelof (see Theorem 6 below). A space $X$ is para-Lindelof if every open cover has a locally-countable open refinement.

Can Theorem 2 hold for point-finite covers (or point-countable covers)? The answer is no (see Example 1 below). With the additional property of having a Baire space, we have the following theorem.

Theorem 3

In any Baire space with the CCC, any point-finite open cover must be countable.

A Space $X$ is a Baire space if $U_1,U_2,U_3,\cdots$ are dense open subsets of $X$, then $\bigcap \limits_{j=1}^\infty U_j \ne \varnothing$. For more information about Baire spaces, see this previous post.
.

Proof of Theorem 3
Let $X$ be a Baire space with the CCC. Let $\mathcal{U}$ be a point-finite open cover of $X$. Suppose that $\mathcal{U}$ is uncountable. We show that this assumption with lead to a contradiction. Thus $\mathcal{U}$ must be countable.

By Theorem 1, there exists an open set $V \subset X$ such that $\mathcal{U}$ is not locally-countable at any point in $V$. For each positive integer $n$, let $H_n$ be the following:

$H_n=\left\{x \in V: x \text{ is in at most } n \text{ sets in } \mathcal{U} \right\}$

Note that $V=\bigcup \limits_{j=1}^\infty H_j$. Furthermore, each $H_n$ is a closed set in the space $V$. Since $X$ is a Baire space, every non-empty open subset of $X$ is of second category (i.e. it cannot be a union of countably many closed and nowhere dense sets). Thus it cannot be that each $H_n$ is nowhere dense in $V$. For some $n$, $H_n$ is not nowhere dense. There must exist some open $W \subset V$ such that $H_n \cap W$ is dense in $W$. Because $H_n$ is closed, $W \subset H_n$.

Choose $y \in W$. The point $y$ is in at most $n$ open sets in $\mathcal{U}$. Let $U_1,U_2,\cdots,U_m \in \mathcal{U}$ such that $y \in \bigcap \limits_{j=1}^m U_j$. Clearly $1 \le m \le n$. Let $U=W \cap U_1 \cap \cdots \cap U_m$. Note that $y \in U \subset H_n \subset V$.

Every point in $U$ belongs to at most $n$ many sets in $\mathcal{U}$ and already belong to $m$ sets in $\mathcal{U}$. So each point in $U$ can belong to at most $n-m$ additional open sets in $\mathcal{U}$. Consider the case $n-m=0$ and the case $n-m>0$. We show that each case leads to a contradiction.

Suppose that $n-m=0$. Then each point of $U$ can only meet $n$ open sets in $\mathcal{U}$, namely $U_1,U_2,\cdots,U_m$. This contradicts that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Suppose that $k=n-m>0$. Let $\mathcal{U}^*=\mathcal{U}-\left\{U_1,\cdots,U_m \right\}$. Let $\mathcal{M}$ be the following collection:

$\mathcal{M}=\left\{U \cap \bigcap \limits_{O \in M} O \ne \varnothing: M \subset \mathcal{U}^* \text{ and } \lvert M \lvert=k \right\}$

Each element of $\mathcal{M}$ is an open subset of $U$ that is the intersection of exactly $n$ many open sets in $\mathcal{U}$. So $\mathcal{M}$ is a collection of pairwise disjoint open sets. The open set $U$ as a topological space has the CCC. So $\mathcal{M}$ is at most countable. Thus the open set $U$ meets at most countably many open sets in $\mathcal{U}$, contradicting that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Both cases $n-m=0$ and $n-m>0$ lead to contradiction. So $\mathcal{U}$ must be countable. The proof to Theorem 3 is completed. $\blacksquare$

As a corollary to Theorem 3, we have the result that every Baire CCC metacompact space is Lindelof.

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Some Applications of Theorems 2 and 3

In proving paracompactness in some of the theorems, we need a theorem involving the concept of star-countable open cover. A collection $\mathcal{A}$ of subsets of a space $X$ is said to be star-finite (star-countable) if for each $A \in \mathcal{A}$, only finitely (countably) many sets in $\mathcal{A}$ meets $A$, i.e., the following set

$\left\{B \in \mathcal{A}: B \cap A \ne \varnothing \right\}$

is finite (countable). The proof of the following theorem can be found in Engleking (see the direction (iv) implies (i) in the proof of Theorem 5.3.10 on page 326 in [1]).

Theorem 4

If every open cover of a regular space $X$ has a star-countable open refinement, then $X$ is paracompact.

As indicated in the above section, Theorem 2 and Theorem 3 have some obvious applications. We have the following theorems.

Theorem 5

Let $X$ be a CCC space. Then $X$ is paracompact if and only of $X$ is Lindelof.

Proof of Theorem 5
The direction $\Longleftarrow$ follows from the fact that any regular Lindelof space is paracompact.

The direction $\Longrightarrow$ follows from Theorem 2. $\blacksquare$

Theorem 6

Every CCC para-Lindelof space is Lindelof.

Proof of Theorem 6
This also follows from Theorem 2. $\blacksquare$

Theorem 7

Every Baire CCC metacompact space is Lindelof.

Proof of Theorem 7
Let $X$ be a Baire CCC metacompact space. Let $\mathcal{U}$ be an open cover of $X$. By metacompactness, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. By Theorem 3, $\mathcal{V}$ must be countable. $\blacksquare$

Theorem 8

Every Baire CCC hereditarily metacompact space is hereditarily Lindelof.

Proof of Theorem 8
Let $X$ be a Baire CCC hereditarily metacompact space. To show that $X$ is hereditarily Lindelof, it suffices to show that every non-empty open subset is Lindelof. Let $Y \subset X$ be open. Then $Y$ has the CCC and is also metacompact. Being a Baire space is hereditary with respect to open subspaces. So $Y$ is a Baire space too. By Theorem 7, $Y$ is Lindelof. $\blacksquare$

Theorem 9

Every locally CCC regular para-Lindelof space is paracompact.

Proof of Theorem 9
A space is locally CCC if every point has an open neighborhood that has the CCC. Let $X$ be a regular space that is locally CCC and para-Lindelof. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the para-Lindelof assumption, let $\mathcal{V}$ be a locally-countable open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a locally-countable open cover. By Theorem 2, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 10

Every locally CCC regular metacompact Baire space is paracompact.

Proof of Theorem 10
Let $X$ be a regular space that is locally CCC and is a metacompact Baire space. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the metacompact assumption, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too. Each open set in $\mathcal{V}$ is also a Baire space.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a point-finite open cover. By Theorem 3, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 11

Every locally compact metacompact perfectly normal space is paracompact.

Proof of Theorem 11
This follows from Theorem 10 after we prove the following two points:

• Any locally compact space is a Baire space.
• Any perfect locally compact space is locally CCC.

To see the first point, let $Y$ be a locally compact space. Let $W_1,W_2,W_3,\cdots$ be dense open sets in $Y$. Let $y \in Y$ and let $W \subset Y$ be open such that $y \in W$ and $\overline{W}$ is compact. We show that $W$ contains a point that belongs to all $W_n$. Let $X_1=W \cap W_1$, which is open and non-empty. Next choose non-empty open $X_2$ such that $\overline{X_2} \subset X_1$ and $X_2 \subset W_2$. Next choose non-empty open $X_3$ such that $\overline{X_3} \subset X_2$ and $X_3 \subset W_3$. Continue in this manner, we have a sequence of open sets $X_1,X_2,X_3,\cdots$ such that for each $n$, $\overline{X_{n+1}} \subset X_n$ and $\overline{X_n}$ is compact. The intersection of all the $X_n$ is non-empty. The points in the intersection must belong to each $W_n$.

To see the second point, let $Y$ be a locally compact space such that every closed set is a $G_\delta$-set. Suppose that $Y$ is not locally CCC at $y \in Y$. Let $U \subset Y$ be open such that $y \in U$ and $\overline{U}$ is compact. Then $U$ must not have the CCC. Let $\left\{U_\alpha: \alpha<\omega_1 \right\}$ be a pairwise disjoint collection of open subsets of $U$. Let $O=\bigcup \limits_{\alpha<\omega_1} U_\alpha$ and let $C=Y-O$.

Let $C=\bigcap \limits_{n=1}^\infty V_n$ where each $V_n$ is open in $Y$ and $V_{n+1} \subset V_n$ for each integer $n$. For each $\alpha<\omega_1$, pick $y_\alpha \in U_\alpha$. For each $y_\alpha$, there is some integer $f(\alpha)$ such that $y_\alpha \notin V_{f(\alpha)}$. So there must exist some integer $n$ such that $A=\left\{y_\alpha: f(\alpha)=n \right\}$ is uncountable.

The set $A$ is an infinite subset of the compact set $\overline{U}$. So $A$ has a limit point, say $p$ (also called cluster point). Clearly $p \notin O$. So $p \in C$. In particular, $p \in V_n$. Then $V_n$ contains some points of $A$. But for any $y_\alpha \in A$, $y_\alpha \notin V_n=V_{f(\alpha)}$, a contradiction. So $Y$ must be locally CCC at each $y \in Y$. $\blacksquare$

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Some Examples

Example 1
A CCC space $X$ with an uncountable point-finite open covers. This example demonstrates that in Theorem 2, locally-finite or locally-countable cannot be replaced by point-finite. Consider the following product space:

$Y=\prod \limits_{\alpha < \omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$

i.e, the product space of $\omega_1$ many copies of the two-point discrete space $\left\{0,1 \right\}$. Let $X$ be the set of all points $h \in Y$ such that $h(\alpha)=1$ for only finitely many $\alpha<\omega_1$.

The product space $Y$ is the product of separable spaces, hence has the CCC. The space $X$ is dense in $Y$. Hence $X$ has the CCC. For each $\alpha<\omega_1$, define $U_\alpha$ as follows:

$U_\alpha=\left\{h \in X: h(\alpha)=1 \right\}$

Then $\left\{U_\alpha:\alpha<\omega_1 \right\}$ is a point-finite open cover of $X$. Of course, $X$ in this example is not a Baire space. $\blacksquare$

The following three examples center around the four properties in Theorem 7 (Baire + CCC + metacompact imply Lindelof). These examples show that each property in the hypothesis is crucial.

Example 2
A separable non-Lindelof space that is a Baire space. This example shows that the metacompact assumption is crucial for Theorem 7.

The example is the Sorgenfrey plane $S \times S$ where $S$ is the real line with the Sorgenfrey topology (generated by the half-open intervals of the form $[a,b)$). It is well known that $S \times S$ is not Lindelof. The Sorgenfrey plane is Baire and is separable (hence CCC). Furthermore, $S \times S$ is not metacompact (if it were, it would be Lindelof by Theorem 7). $\blacksquare$

Example 3
A non-Lindelof metacompact Baire space $M$. This example shows that the CCC assumption in Theorem 7 is necessary.

This space $M$ is the subspace of Bing’s Example G that has finite support (defined and discussed in the post A subspace of Bing’s example G. It is normal and not collectionwise normal (hence cannot be Lindelof) and metacompact. The space $M$ does not have CCC since it has uncountably many isolated points. Any space with a dense set of isolated points is a Baire space. Thus the space $M$ is also a Baire space. $\blacksquare$

Example 4
A non-Lindelof CCC metacompact non-Baire space $W$. This example shows that the Baire space assumption in Theorem 7 is necessary.

Let $W$ be the set of all non-empty finite subsets of the real line with the Pixley-Roy topology. Note that $W$ is non-Lindelof and has the CCC and is metacompact. Of course it is not Baire. For more information on Pixley-Roy spaces, see the post called Pixley-Roy hyperspaces. For the purpose of this example, the Pixley-Roy space can be built on any uncountable separable metrizable space.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Tall, F. D., The Countable Chain Condition Versus Separability – Applications of Martin’s Axiom, Gen. Top. Appl., 4, 315-339, 1974.

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$\copyright \ 2014 \text{ by Dan Ma}$